Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the BST.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q); if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q); return root; } }这个解法简直钛蚌
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先看recursive。从根节点出发，分别去搜索两个节点，但这里搜索不用瞎搜索，需要利用BST的性质，比较p->val，q->val和root->val的关系，决定往左边搜还是右边搜。如果root->val>p->val同时root->val>q->val,说明p,q都在root的左侧，因此把root=root->left;如果root->val< p->val同时root->val < q->val,说明p,q都在root的右侧，因此把root=root->right;如果root->val>p->val同时root->val< q->val,或root->val< p->val同时root->val> q->val说明p,q在root的两侧，说明root就是LCA.
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